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If one number is thrice the other and their sum is $16$, find the numbers.

I tried, Let the first number be $x$ and the second number be $y$ Acc. to question

$$ \begin{align} x&=3y &\iff x-3y=0 &&(1)\\ x&=16-3y&&&(2) \end{align} $$

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  • $\begingroup$ One number is $x$, the other $3x$. They add to 16, so ... . $\endgroup$ – Chris Leary Jul 27 '14 at 20:01
  • $\begingroup$ the second part of the questions states "their sum is 16". this can be expressed as x+y=16 $\endgroup$ – Mufasa Jul 27 '14 at 20:01
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The problem statement says $$x+3x=16,$$ hence $$x=4,\\3x=12.$$

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I assume you have $$x=\color{red}{3y}, ~~x+y=16$$ Then $3(x+\color{red}{y})=3\times 16=48$ and so $3x+\color{red}{3y}=48$ and so $3x+x=48$ and so $4x=48$...

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$$x=3y$$ $$y+x=16 \Rightarrow y+3y=16 \Rightarrow 4y=16 \Rightarrow y=4$$

So, $x=12$.

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Let the 1st number be $x$ and the 2nd number be $3x$. Since $x + 3x = 16$, $4x = 16$, so $x=4$. Therefore, 1st number is $4$ and the other is $12$.

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Let the first number be $x$.

Let the second number be $y$.

According to question

$$ \tag{1} x+y=16 $$ $$ \tag{2} x=3y $$ So, $x-3y=0 \tag{2}$ Multiply equation $(1)$ by $3$.

Solve both equations:

$$\tag{1} 3x+3y=48$$ $$\tag{2} x-3y=0$$ $$\tag{1) + (2}4x=48$$ $$\tag{3}x=12$$ Putting in equation $(1)$: $$\tag{1} x+y=16$$ $$\tag{1),(3} 12+y=16$$ $$\tag{4}y=16-12$$ $$\tag{5}y=4$$

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    $\begingroup$ Welcome to MSE! Please use MathJax so the others can better benefit from your questions and answers. $\endgroup$ – Bill O'Haran May 3 '18 at 15:57
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Let the first number be $x$ And second which is thrice be $3y$

Acc to ques..

$x=3y\ldots \textrm{equation 1}$

$x+y=16\ldots\textrm{equation 2}$

By Elimination method

       x-3y  =    0
     -(x+ y) =  -16
     ______________
       0- 4y = - 16

Dividing the equation by $(-4)$. The result is $y=4$

Put $y=4$ into the first equation :$x=3\cdot 4\Rightarrow x=12$

I hope this will help you.. Thank you! Please like my comment if you like it

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